Problems for Chapter 1 & 2 ANSWERS
1. Draw an appropriate Lewis electron dot structure for the compound HONNH.
Valence electrons
2xH = 2
2xN = 10
O =6
TOTAL = 18
H N
N O H
2. Show the formal charge, if one exists, on each atom in the Lewis electron dot
structure at below:
FC = 6-!/2(4) - 4 = O
O
FC = 4-!/2(8) - 0 = O
C
HO
O
FC = 6-!/2(2) - 6 = -1
FC = 6-!/2(4) - 4 = )
3. Rewrite the structural formula, below, using a Kekule structure.
(CH3)2CHCH(CH3)CH=CHCH(OH)CH3
H CH 3
H3C C C C C
CH 3H H H
O H
C CH 3
H
4. Show the three-dimensional structure for the compound, AlBr3. If the
compound has an overall dipole moment, show the direction of that dipole with
an arrow, +→.
Br
Al
Br
Br
These two dipoles, have vectors which resolve into a
vector equal in magnitude to, but opposite direction to
the one pointing up...NO NET DIPOLE
Trigonal Planar Shape
(flat, like a 3 bladed
propellar)
5. Show the three-dimensional structure for the compound, NH3. If the compound
has an overall dipole moment, show the direction of that dipole with an arrow,
+→.
Pyramidal shape,
based on sp3
nitrogen
NET DIPOLE
N
H
H
H
6. Draw a Lewis dot structure for the following compounds, show formal charge
if any
a. CH3CH=O b. NH4+
H
H C C
H H
H +1
H N H
H
O
7. Draw the Kekule structures for the above condensed structures.
H
H
H C
C
H
H
O
H
N
H
H
8. Draw the 3-D shape of each molecule, in question (6) above, about the
a. right carbon atom
b. nitrogen
H
CH 3
C
O
H
Trigonal Planar
H
H
N
H
Tetrahedral
9. Name the 3-D shape of each molecule in question (6), about the:
a. right carbon atom
b. nitrogen
Trigonal Planar
Tetrahedral
10. Name the orbital hybridization for the compounds in question (6) above, about
the:
a. right carbon atom
b. nitrogen
2
sp
sp 3
11. What will be the approximate bond angles in the molecules in question (6)
above, about the:
a. right carbon atom
b. nitrogen
120˚
109˚
12. Carbon can exist in a trivalent state (three bonds) but neutral in charge. Predict
(show) the Lewis structure for such a carbon with all bonds to hydrogen.
H
H
C
H
13.
What orbital hybridization is likely for the carbon in the above species?
Remember that hybridized electrons are always paired (bonded or nonbonded). Draw the carbon showing all orbitals.
H
C
H Carbon is sp 2 hybridized, with one p
orbital left for the unpaired electron
H to reside in
14.Draw the Kekulé structure for the skeletal drawing shown below.
CH 3
H H CH 3 H CH 3 H
H C C
C
C
C
H H
H
CH 3 H
H
C H
H
15.Draw the above compound in a condensed structure.
CH 3CH 2C[CH(CH 3)2]C(CH 3)2CH 2CH 3
16. Show, with an arrow, the polar bonds and the direction of polarity in the
following molecules.
H
H H O H
Cl
C C C C H
H H CH3 H
H H O H
H
H C C C C N
H H
H
H
17. What will be the numerical value of the Ka whose pKa is 4.2?
(You may use two different powers of ten as your answer.)
pKa = - Log Ka,
pKa = 4.2
Ka = 10 -4.2 = 0.00063
18 If a compound is 62.041% carbon and 10.412% hydrogen, what is its empirical
formula?
If there is 62.031% carbon and 10.412% hydrogen, there must be 27.547% oxygen
in sample.
In 100 g of sample there is: 62.041 g C, 10.412 g H and 27.547 g O.
In 100 g of sample there is: 62.041g/12.011 = 5.16 moles of Carbon,
10.412g/1.008 = 10.33 moles of Hydrogen
27.547g/15.999 = 1.72 moles of Oxygen
Molar ratio :C5.16 , H10.33 , O1.72, divide all by smallest ratio i.e. 1.72, formula becomes
C 3H6O
19. Identify the acids and bases in the following equation.
O
O
CH 3CH2 CH2 C–H +
CH 3CH2 CH2CH 2–
acid
+
CH3CH 2CHC–H + CH3 CH2 CH2CH 3
Li
base
conjugate base conjugate acid
20. The OH ion is the conjugate base of H2O. If water has a pKa of 15.7, which of
the following acids would be completely neutralized by NaOH? Circle them.
(Bold is correct answer)
Acid
A
B
C
pKa
14
20
33
Keq = 10 Pka(conj acid)
-
A-H +
pKa =14
Keq
-
OH
B-H +
pKa =20
-
OH
C-H +
pKa =33
-
OH
Keq
Keq
10
pKa (acid)
A
+ H2O
Keq = 101.7
pKa = 15.7
B-
+ H2O
Keq = 10-4.3
pKa = 15.7
C-
Keq = 10-17.3
+ H2O
pKa = 15.7
21. Draw the following molecules in their three dimensional form and show, on the
structure, the following: (1) The polarity of each polar bond (—>) and (2) The
overall polarity direction of the molecule (+——>). If there is no molecular
dipole moment, circle the structure
CH3 NH 2
H
H
N
C
H
H
BF3
F
Net Dipole
Dipole moments
Cancell
B
F
F
H
22. Compounds ‘A’ and ‘B’, at right, have the same polar C–N bond. However,
‘A’ has a dipole moment of 1.31 D while that of ‘B’ is 2.11 D. Explain why ‘B’
has almost double the dipole moment of ‘A’. Illustrate your explination.
(Hint: Look at formal charges in structure B)
H
H
N
C
H
H
H
Net Dipole
H
H C
H
N
N
N
Extremely large dipole
oriented toward negative charge
23. In each of the following chemical reactions, circle the acid and place an ‘X’
over the conjugate base.
X
+–
O
O
CH3CCH2CCH3 +
–
NH2
AlCl 3 + HCl
–+
CH3CH2O
X
O
CH3
Na + CH4
X
CH3CH2OH + Na
O
–
CH3CCHCCH3 + NH3
–
+
AlCl 4 + H
+ –
AlCl 3 + NH3
Lewis Acid
H3N – AlCl 3
24. Combustion analysis of an organic compound shows it to have 64.3% carbon
and 7.2% hydrogen. What is the emperical formula for the compound?
If there is 64.3% carbon and 7.2% hydrogen, there must be 28.5% oxygen in sample.
In 100 g of sample there is: 64.3 g C, 7.2 g H and 28.5 g O.
In 100 g of sample there is: 64.3g/12.011 = 5.35 moles of Carbon,
7.2g/1.008 = 7.14 moles of Hydrogen
28.5g/15.999 = 1.78 moles of Oxygen
Molar ratio :C5.35 , H7.14 , O1.78, divide all by smallest ratio i.e. 1.78, formula becomes C3.00 H4.01 O
C 3H4O
25.Identify the Lewis acids and bases in the reactions of question (23).
Lewis acids are electron acceptors
Lewis bbases are electron donors
+–
–+
CH3CH2OH + Na CH3
Lewis Acid Lewis Base
O
O
CH3CH2O
O
–
NH2
Na + CH4
O
–
CH3CCH2CCH3 +
Lewis Acid Lewis Base
CH3CCHCCH3 + NH3
–
+
AlCl 3 + HCl
Lewis Acid Lewis Base
AlCl 4 + H
AlCl 3 + NH3
Lewis Acid Lewis Base
H3N – AlCl 3
+ –
26.How can you explain the fact that the O-H bond in methanol(CH3-OH) is more
acidic than the C-H bonds.
Compare the conjugate bases which are formed when methanol (CH3-OH) acts
as an acid by breaking either the O-H or C-H bond:
CH 3O H
CH 3O + H +
Negative charge resides
on more electronegative
oxygen atom.
HOH 2C H
HOH 2C + H+
Negative charge resides
on less electronegative
carbon atom.
The conjugate base with the negative charge on the oxygen is more stable